Wavelenght of the incident light

  • Thread starter Thread starter wyclefchick
  • Start date Start date
  • Tags Tags
    Light Wavelenght
AI Thread Summary
The discussion revolves around calculating the wavelength of incident light on a mercury plate with a known work function and stopping voltage. The participant initially attempts to use the equation involving Planck's constant but encounters discrepancies in their calculations. They mistakenly use an incorrect value for Planck's constant, leading to an incorrect wavelength result. Clarifications are made regarding the proper value of Planck's constant and the correct calculations needed to arrive at the expected wavelength of 201.7561 nm. The conversation emphasizes the importance of using accurate constants and following the correct mathematical steps.
wyclefchick
Messages
9
Reaction score
0

Homework Statement


An incident light shines on a mercury metal plate with a work function of -4.5 eV. The stopping voltage is observed to be -1.65 volts. What is the wavelenght of the incident light?


Homework Equations


wavelength=hc/E-0(it's a Greek sign n i don't have it in my computer)


The Attempt at a Solution


(4.14 X 10^-15 eVs) (3 X 10^8 m/s) / 1.65 + 4.5 = 1.242 X 10^-6 / 6.15

1.242(10^2)/6.15 = 20.1954

but the right answer is 201.7561 nm


so can anyone please tell me what did I do wrong?
 
Physics news on Phys.org
\frac{hc}{\lambda}=\phi +eV_s

Did you use the correct value for h in eV?
 
i used h=4.14 X 10^ -15 am i supposed to use the other one? h=6.626 X 10^-34 ?
 
How'd you go from 1.242 X 10^-6 / 6.15 to 1.242(10^2)/6.15 = 20.1954?

Because if you work out 1.242 X 10^-6 / 6.15 you'd get your answer.
 
I was following one of the examples that my teacher did in class.

well if i do 1.242 X 10^-6 / 6.15...it equals 2.01951 but the right answer is 201.7561 nm
 
1.242 X 10^-6 / 6.15 =201nm
 
Kindly see the attached pdf. My attempt to solve it, is in it. I'm wondering if my solution is right. My idea is this: At any point of time, the ball may be assumed to be at an incline which is at an angle of θ(kindly see both the pics in the pdf file). The value of θ will continuously change and so will the value of friction. I'm not able to figure out, why my solution is wrong, if it is wrong .
TL;DR Summary: I came across this question from a Sri Lankan A-level textbook. Question - An ice cube with a length of 10 cm is immersed in water at 0 °C. An observer observes the ice cube from the water, and it seems to be 7.75 cm long. If the refractive index of water is 4/3, find the height of the ice cube immersed in the water. I could not understand how the apparent height of the ice cube in the water depends on the height of the ice cube immersed in the water. Does anyone have an...
Back
Top